What would be the temperature inside the pressure cooker if the vapor pressure of water was 3.60 atm? The enthalpy of vaporization of water is 40.7 kJ/mol.

Answer 1

Clausius-Clapeyron equation:

ln (P2/P1) = -∆Hvap/R (1/T2 - 1/T1)

P1 = 1 atm

P2 = 3.60 atm

∆Hvap = 40.7 kJ/mol

R = 8.314 J/Kmol = 0.008314 kJ/Kmol

T1 = 100ºC = 373K

T2 = ?

ln (3.60/1.0) = -40.7/0.008314 (1/373 - 1/T2)

1.28 = -4895 (1/373 - 1/T2)

1.28 = -13.1 + 4895/T2

14.38 = 4895/T2

T2 = 340K = 67ºC